結果
問題 | No.3098 tan 2 |
ユーザー | omgflaka |
提出日時 | 2023-03-31 22:00:28 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 2,538 bytes |
コンパイル時間 | 3,156 ms |
コンパイル使用メモリ | 218,712 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-09-23 00:40:33 |
合計ジャッジ時間 | 3,692 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
6,812 KB |
testcase_01 | WA | - |
testcase_02 | WA | - |
testcase_03 | WA | - |
testcase_04 | WA | - |
testcase_05 | WA | - |
testcase_06 | WA | - |
testcase_07 | WA | - |
testcase_08 | WA | - |
testcase_09 | WA | - |
testcase_10 | WA | - |
testcase_11 | WA | - |
ソースコード
#pragma GCC optimize("O3,Ofast,unroll-loops") #pragma GCC target("avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #pragma comment(linker, "/stack:20000000") #include <bits/stdc++.h> #define _CRT_SECURE_NO_WARNINGS using namespace std; typedef long long ll; #define vi vector<int> #define vb vector<bool> #define pub push_back #define emb emplace_back #define rep(i,a,b) for(long long i = a; i < b; i++) #define rrep(i,a,b) for(long long i = a; i >= b; i--) #define iterm(key,val,ok) for(auto const& [key, val]:ok) #define all(x) x.begin(),x.end() #define py cout<<"YES"<<endl #define pn cout<<"NO"<<endl #define pa cout<<ans<<endl #define pe cout<<endl #define mp make_pair #define f first #define se second using pii=pair<ll,ll>; template<typename T> istream& operator>>(istream& is, vector<T>& v) { for (auto& i : v) is >> i; return is; } template<typename T> ostream& operator<<(ostream& os, const vector<T>& v) { for (auto& i : v) os << i << " "; return os; } template<typename T> istream& operator>>(istream& is, pair<T, T>& v) { is >> v.first >> v.second; return is; } template<typename T> ostream& operator<<(ostream& os, const pair<T, T>& v) { os << v.first << " " << v.second; return os; } /* problems with mod or small constraints are DP (2d,3d) or combo ascii: 1=49 a=97 A=65 check edge cases and constraints CHECK FOR JUST ONE ELEMENT OR SAME NUMBERS ETC always think about reversing a process see if theres any dumb tricks if youre not getting it: like only having to check small values or sth (a-b)modm = (a%m+(m-b%m))%m mark visited after adding to queue think about transforming representations of graphs, if some graph is too big to be represented, think about bipartite representation, if its too small maybe you can expand according to problem to make it easier you can also bfs on pairs of nodes, different representations of nodes etc etc lower bound can give beyond end also (check segfaults etc if using) many smol vectors are slowwww, use std:array instead to search for and remove particular things fastly just use multiset, like u can use prefix multiset and find closest value of something also like so many applications in just logn time */ void solve(){ ll n;cin>>n; if(n==45){ cout<<"Yes"<<endl; } else{ cout<<"No"<<endl; } } signed main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int t=1; cin>>t; while(t--){ solve(); } }